2024 AMC 12A Problems/Problem 23

Problem

What is the value of

Solution 1 (Trigonometric Identities)

First, notice that \begin{align*} &\tan^2 \frac{\pi}{16} \cdot \tan^2 \frac{3\pi}{16} + \tan^2 \frac{\pi}{16} \cdot \tan^2 \frac{5\pi}{16} + \tan^2 \frac{3\pi}{16} \cdot \tan^2 \frac{7\pi}{16} + \tan^2 \frac{5\pi}{16} \cdot \tan^2 \frac{7\pi}{16} \\ &= \left( \tan^2 \frac{\pi}{16} + \tan^2 \frac{7\pi}{16} \right) \cdot \left( \tan^2 \frac{3\pi}{16} + \tan^2 \frac{5\pi}{16} \right) \end{align*}

Here, we use the identity \begin{align*} \tan^2 x + \tan^2 \left( \frac{\pi}{2} - x \right) &= \left( \tan x + \tan \left( \frac{\pi}{2} - x \right) \right)^2 - 2 \\ &= \left( \frac{\sin x}{\cos x} + \frac{\sin \left( \frac{\pi}{2} - x \right)}{\cos \left( \frac{\pi}{2} - x \right)} \right)^2 - 2 \\ &= \left( \frac{\sin x \cos \left( \frac{\pi}{2} - x \right) + \sin \left( \frac{\pi}{2} - x \right) \cos x}{\cos x \cos \left( \frac{\pi}{2} - x \right)} \right)^2 - 2 \\ &= \left( \frac{\sin \frac{\pi}{2}}{\cos x \cos \left( \frac{\pi}{2} - x \right)} \right)^2 - 2 \\ &= \left( \frac{1}{\cos x \sin x} \right)^2 - 2 \\ &= \left( \frac{2}{\sin 2x} \right)^2 - 2 \\ &= \frac{4}{\sin^2 2x} - 2 \end{align*}

Hence, \begin{align*} \left( \tan^2 \frac{\pi}{16} + \tan^2 \frac{7\pi}{16} \right) \cdot \left( \tan^2 \frac{3\pi}{16} + \tan^2 \frac{5\pi}{16} \right) &= \left( \frac{4}{\sin^2 \frac{\pi}{8}} - 2 \right) \left( \frac{4}{\sin^2 \frac{3\pi}{8}} - 2 \right) \end{align*}

Note that \begin{align*} \sin^2 \frac{\pi}{8} &= \frac{1 - \cos \frac{\pi}{4}}{2} = \frac{2 - \sqrt{2}}{4} \\ \sin^2 \frac{3\pi}{8} &= \frac{1 - \cos \frac{3\pi}{4}}{2} = \frac{2 + \sqrt{2}}{4} \end{align*}

Thus, \begin{align*} \left( \frac{4}{\sin^2 \frac{\pi}{8}} - 2 \right) \left( \frac{4}{\sin^2 \frac{3\pi}{8}} - 2 \right) &= \left( \frac{16}{2 - \sqrt{2}} - 2 \right) \left( \frac{16}{2 + \sqrt{2}} - 2 \right) \\ &= (14 + 8\sqrt{2})(14 - 8\sqrt{2}) \\ &= 68 \end{align*}

Therefore, the answer is \(\boxed{\textbf{(B) } 68}\).

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.